Question

Q1/ A temperature rise of 60°C in a circular shaft of 60 mm diameter is caused by the amount of heat generated. The thermal conductivity of the shaft material is 50 W/m.°C and the heat transfer coefficient is 6.5 W/m². C. 1. Develop an expression for the temperature distribution; 2. Determine the heat transfer from the shaft. Assume the shaft is a rod of infinite length.

Answer 1

Temperatusie ruse = 60°C diameter d= Gomm ka 50cm'c h= 6. 50 L=0 Infinite length 0 Temperature Distribution © Heat Trawiferi zoc x Solver Hot surface I to EN Tott FL- + OLD) < Txp+XO Given (Laos tų tot Let Toy Root or Bane Temperature Tood Ambient temperature AT = (To-To) - Temperature rise by Convective heat Trauffer coff. Ko Thermal Conductivity do dia al shaft Le length of fin to

ŕ shaft - let in a Y is heat direction Conducted Conduce qux = -K Ac dxd fourier law "Ac d Tobu 1. = Acy of heat Conduction of" front sectional area t let dx aree 402 & axedy is the sa elment t a q xfax - heat Conducted ñ Xo. » arx+dx = ax & J V (2x) ax queted ñ xodiore etian û Element - By Newton's law of cooling convected = b As (T- Two) (3) T = Temperature of Element 2 As = surface area o As= perimeter x de & Aso r d xdx * Now Heat Balance 1. Qx = Ox +d) + Q convected - ax = axtax t & conue - ax = orx to evx) dx + Qconui - x-qx = 8 (-KACJdx f RA (T-TO)

90 = o C - KAS 97) dx + (Pdes (Toto) - - Polo C1-P) = 9 CT-poo) 2 Ldx² =0 hp KAC og let 10 = (T-Tos) ]mben KAC re NH 22 o dar2 , et on my Boo] o after solve this

After solve this - O= çe +mx + ca e-mx 710=ci e mx + ca e-my saution of ean ®

Ric » Boundary Conditions & I at x=0 31 at x 100 0:0 T-To= To-Tool IT- Tool Oo= To - Too 0-0 * Apply Bic. 2) into ean 6 > G-07 at X=L=60, 7-10-0 - 10 = C₂ e-mx 7 0 Apply Bc. J into ean & eat x=0, [To-To = do] [Co=oo) 3 [0o=C2] Pinally = 0 = Cq e-mx +10= do e-my

e-mx I T- TAS to Tose-my] - ® T 1 » S Temperature distribution equation T = Temperature after distance xor tip temperature almahe I RAC I E 2 22222222222nnnno 2 Heat Trauffer & CQ = Q = uk Ac ( 1 ) x = 0 Q=ok Ac c-e-mx m] oo Q=tk Ac moo =Q=kac ſhe do Q= JK A c hp oo - Q v AP KAC (Tortos /

- = d = To Give problem 1то - Tom = 60 °C b. 6, 5 w/ м3 °C Gomm К= 50 м'е - P= = = о 1800ч94 - Ас - па? - 2. рэхо-2 - - ув'SX 50х о 1604 x 2.829xio-3 хво : ~ [Q = очi uz | - | Q |о- ні о 1